Theorem 2.5 Rudin functional analysis, use of balanced neighbors.

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Related to this question (I sometimes read through the book to refresh my background in functional analysis).

The theorem starts with picking balanced neighbours $U$, $W$ such that $$ \bar{U} + \bar{U} \subset W $$

But the conclusion of the theorem is reached through the chains of inclusions

$$ \Lambda(V) \subset \Lambda x - \Lambda (E) \subset \bar{U} - \bar{U} \subset W $$

I'm comparing $\bar{U} + \bar{U} \subset W$ against $\bar{U} - \bar{U} \subset W$. In chapter 1 the definition of balanced is:

A set $B \subset X$ (topological vector space) is said to be balanced if $\alpha B \subset B$ for every $\alpha \in \Phi$ with $|\alpha| \leq 1$.

Therefore I suppose we have the inclusion $- \bar{U} \subset \bar{U}$ (since $|-1|\leq 1$) and therefore

$$ \bar{U} - \bar{U} \subset \bar{U} + \bar{U} $$

However my question is whether or not is $- \bar{U} \subset \bar{U}$ or $- \bar{U} = \bar{U}$, I think the latter is the definition of symmetric set.

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$-B \subset B$ if $B$ is balanced. This actually implies $-B=B$. Now $x \to -x$ is a homeomorphism so $- (\overline {U})=\overline {(-U)}=\overline {U}$.